jarkko kari- the tiling problem revisited

The tiling problem revisited

Jarkko Kari

Mathematics Department, University of Turku, Finland

Consider the following decision problem, the tiling problem:

Given a finite set of tiles (say, for example, polygons), is it possible

to tile the infinite plane with copies of the tiles ?

For instance, can one tile the plane with copies of

By a tiling we mean a covering of the plane without overlaps (i.e.

two tiles may only overlap in their boundary).

But there are other tile sets that do not admit any tiling, e.g. the

regular pentagon

does not tile the plane.

So is there an algorithm to tell which tile sets admit a plane tiling ?

So is there an algorithm to tell which tile sets admit a plane tiling ?

R.Berger 1966: No, the tiling problem is undecidable.

R.Berger: Undecidability of the Domino Problem. Memoirs

of the American Mathematical Society 66, 72 pp., 1966.

A simplified version (but based on Berger’s ideas) was given by

R.M.Robinson in 1971.

R.M.Robinson. Undecidability and nonperiodicity fortilings of the plane. Inventiones Mathematicae 12, 177–209,1971.

In this talk we present a new, quite different proof.

Why a new proof to an old result ?



• The new proof is simpler. It is based on simple algebra, it is

precise and easy to verify.



• The new proof is simpler. It is based on simple algebra, it is

precise and easy to verify.

• The same technique works in other set-ups as well. In particular,

the same approach shows that the tiling problem on the hyperbolic

plane is undecidable. This is an open problem posed in Robinson’s

1971 paper, and investigated later in more depth in

R.M.Robinson: Undecidable tiling problems in thehyperbolic plane. Inventiones Mathematicae 44, 259–264,1978.

Our proof is based on constructing tiles that simulate certain

dynamical systems. The dynamical systems that we use are

piecewise affine transformations of Rd. The construction can be

done for any d, but d = 2 is sufficient to prove undecidability.

The undecidability of the tiling problem will be concluded from the

undecidability of the immortality problem of these dynamical

systems.

Outline of the talk

• Introduction. Historical perspective.

• Wang tiles

• The immortality problem of piecewise affine transformations.

• Reductions:

(a) Immortality problem of Turing machines −→ immortality

problem of piecewise affine transformations of R2.

(b) Immortality problem of piecewise affine maps −→ the tiling

problem

• Undecidability of the tiling problem in the hyperbolic plane.

Wang tiles

A Wang tile is a unit square tile with colored edges. A tile set T is

a finite collection of such tiles. A valid tiling is an assignment

Z2 −→ T

of tiles on infinite square lattice so that the abutting edges of

adjacent tiles have the same color.

Wang tiles

A Wang tile is a unit square tile with colored edges. A tile set T is

a finite collection of such tiles. A valid tiling is an assignment

Z2 −→ T

of tiles on infinite square lattice so that the abutting edges of

adjacent tiles have the same color.

For example, consider Wang tiles

A B C D

With copies of the given four tiles we can properly tile a 5× 5square. . .

A

B

C

D

C

A

C

B

D

C

B

D

A

C

C

B

D

C

A

C

B

A

C

D

C

. . . and since the colors on the borders match this square can berepeated to form a valid periodic tiling of the plane.

The tiling problem of Wang tiles is the decision problem to

determine if a given finite set of Wang tiles admits a valid tiling of

the plane.

Theorem (R.Berger 1966): The tiling problem of Wang tiles is

undecidable.

Note: Wang tiles are abstract tiles, but one can effective transform

them into equivalent concrete shapes (e.g. polygons with rational

coordinates).

For example, we can make each Wang tile into a unit square tile

whose left and upper edges have a bump and the right and lower

edge has a dent. The shape of the bump/dent depends on the color

of the edge. Each color has a unique shape associated with it (and

different shapes are used for horizontal and vertical colors).

A B C D

DCBA

Another note: In his proof Berger constructed an aperiodic tile

set. This is a tile set that tiles the plane, but it does not admit any

periodic tiling. This refuted a conjecture by Hao Wang that all tile

sets that admit a tiling admit also a periodic one.

Another note: In his proof Berger constructed an aperiodic tile

set. This is a tile set that tiles the plane, but it does not admit any

periodic tiling. This refuted a conjecture by Hao Wang that all tile

sets that admit a tiling admit also a periodic one.

If Wang’s conjecture had been true then the tiling problem would

be decidable: One could try all possible tilings of larger and larger

rectangles until either

(a) a rectangle is found that can not be tiled (so no tiling of the

plane exists), or

(b) a tiling of a rectangle is found such that the colors at left and

right sides match and the colors of the top and bottom sides

match each other (so a periodic tiling exists).

Only aperiodic tile sets fail to reach either (a) or (b). . .

Immortality of piecewise affine maps

Consider a system of finitely many pairs (Ui, fi) where

• Ui are disjoint unit squares of the plane with integer corners,

• fi are affine transformations with rational coefficients.

Square Ui is understood as the domain where fi may be applied.

The system determines a function

f : D −→ R2

whose domain isD =

⋃

i

Ui

andf(~x) = fi(~x) for all ~x ∈ Ui.

The orbit of ~x ∈ D is the iteration of f starting at point ~x. The

iteration can be continued as long as the point remains in the

domain D.

But if the point goes outside of the domain, the system halts.

If the iteration always halts, regardless of the starting point ~x, the

system is mortal. Otherwise it is immortal: there is an immortal

point ~x ∈ D from which a non-halting orbit begins.

Immortality problem: Is a given system of affine mapsimmortal?

Proposition: The immortality problem is undecidable.

To prove the undecidability one can use a standard technique fortransforming Turing machines into two-dimensional piecewiseaffine transformations.

Turing machine configuration

a b c d e f g h i

q

is encoded as the pair (x, y) ∈ R2 where the digits of x and y (insome suitably large base B) express the contents of the left andright halves of the tape:

x = ef.ghi . . .

y = qd.cba . . .

The integer parts of x and y determine the next move of themachine, that is, the next move depends on the integer unit squarecontaining point (x, y).

a b c d f g h i

q r

xe

A left move of the Turing machine requires that the digits of x andy are shifted one position to the right and left, respectively. Addingsuitable (integer) constants takes care of changes in the state q andthe current tape symbol e.

x = ef.ghi . . .

y = qd.cba . . .7→

x′ = dx.fghi . . .

y′ = rc.ba . . .

This is an affine transformation whose matrix is

1B 0

0 B

.

a b c d f g h i

q r

xe

Analogously, a right move is simulated by an affine transformation

whose matrix is B 0

0 1B

.

Additional changes in the integer parts complete the

transformation:

x = ef.ghi . . .

y = qd.cba . . .7→

x′ = fg.hi . . .

y′ = rx.dcba . . .

A given Turing machine is converted in this way into a system of

unit squares Ui and corresponding affine transformations fi. Then

iterations of the Turing machine on arbitrary configurations

correspond to iterations of the affine maps.





In particular, the system of affine maps has an immortal point if

and only if the Turing machine has an immortal configuration,that is, a configuration that leads to a non-halting computation in

the Turing machine. But we have the following result:

Theorem (Hooper 1966): It is undecidable if a given Turing

machine has any immortal configurations.





In particular, the system of affine maps has an immortal point if

and only if the Turing machine has an immortal configuration,that is, a configuration that leads to a non-halting computation in

the Turing machine. But we have the following result:

Theorem (Hooper 1966): It is undecidable if a given Turing

machine has any immortal configurations.

(Interesting historical note: Hooper and Berger were both students

of Hao Wang, at the same time. Their results are of same flavor

but the proofs are independent.)

Immortality problem: Is a given system of affine maps

immortal?

Proposition: The immortality problem is undecidable.

The proposition now follows from Hooper’s theorem.

Next we reduce the immortality problem to the tiling problem. The

idea is the same as in my earlier construction of 14 aperiodic Wang

tiles: In that construction the tiles simulated the piecewise linear

function f : [ 12 , 2] −→ [ 1

2 , 2] where

f(x) =

2x, if x ≤ 1, and23x, if x > 1.

Iterations of f are non-periodic, from which aperiodicity of the

corresponding tile set follows.

Next we reduce the immortality problem to the tiling problem. The

idea is the same as in my earlier construction of 14 aperiodic Wang

tiles: In that construction the tiles simulated the piecewise linear

function f : [ 12 , 2] −→ [ 1

2 , 2] where

f(x) =

2x, if x ≤ 1, and23x, if x > 1.

Iterations of f are non-periodic, from which aperiodicity of the

corresponding tile set follows.

That construction is now generalized

• from linear maps to affine maps, and

• from R to R2, (or Rd for any d).

The colors in our Wang tiles are elements of R2.

Let f : R2 −→ R2 be an affine function. We say that tile

n

w

s

e

computes function f if

f(~n) + ~w = ~s + ~e.

(The ”input” ~n comes from north, and f(~n) is computed. A ”carry

in” ~w from the west is added, and the result is split between the

”output” ~s to the south and the ”carry out” ~e to the east.)

Suppose we have a correctly tiled horizontal segment of length n

where all tiles compute the same f .

Average =

e

s

n

w

Average =

It easily follows that

f(~n) +1n

~w = ~s +1n

~e,

where ~n and ~s are the averages of the top and the bottom labels.

Suppose we have a correctly tiled horizontal segment of length n

where all tiles compute the same f .

Average =

e

s

n

w

Average =

It easily follows that

f(~n) +1n

~w = ~s +1n

~e,


As the segment is made longer, the effect of the carry in and outlabels ~w and ~e vanish. In the limit, if we have an infinite row oftiles, the average of the input labels (if it exists!) is mapped by f

to the average of the output labels.

Consider a system of affine maps fi and unit squares Ui.

For each i we construct a set Ti of Wang tiles

• that compute function fi, and

• whose top edge labels ~n are in Ui.

An additional label i on the vertical edges makes sure that tiles ofdifferent sets Ti and Tj cannot be mixed on any horizontal row oftiles. Let

T =⋃

i

Ti.

If T admits a valid tiling then the system of affine maps has an

immortal point:

• Consider any horizontal row in a valid tiling. The top labels

belong to a compact and convex set Ui. Hence there is ~x ∈ Ui

that is the limit of the top label averages over a sequence of

segments of increasing length.


immortal point:





• Then fi(~x) is the limit of the bottom label averages over the

same sequence of segments.


immortal point:







• But the bottom labels of a row are the same as the top labels

of the next row below, so fi(~x) is the limit of the top label

averages of the next row.


immortal point:







• But the bottom labels of a row are the same as the top labels

of the next row below, so fi(~x) is the limit of the top label

averages of the next row.

• The reasoning is repeated for the next row, and for all rows

below.

If T admits a valid tiling then the system of affine maps has animmortal point:

• Consider any horizontal row in a valid tiling. The top labelsbelong to a compact and convex set Ui. Hence there is ~x ∈ Ui

that is the limit of the top label averages over a sequence ofsegments of increasing length.

• Then fi(~x) is the limit of the bottom label averages over thesame sequence of segments.

• But the bottom labels of a row are the same as the top labelsof the next row below, so fi(~x) is the limit of the top labelaverages of the next row.

• The reasoning is repeated for the next row, and for all rowsbelow.

• We see that ~x starts an infinite orbit of the affine maps, so it isimmortal.

We still have to detail how to choose the tiles so that also the

converse is true: any immortal orbit of the affine maps corresponds

to a valid tiling.

Consider a unit square

U = [n, n + 1]× [m,m + 1]

where n,m ∈ Z. Elements of

Cor(U) = {(n,m), (n,m + 1), (n + 1,m), (n + 1,m + 1)}

are the corners of U .

For any ~x ∈ R2 and k ∈ Z denote

Ak(~x) = bk~xc

where the floor is taken for each coordinate separately:

b(x, y)c = (bxc, byc).

Denote

Bk(~x) = Ak(~x)−Ak−1(~x) = bk~xc − b(k − 1)~xc.

It easily follows that if ~x ∈ U then

Bk(~x) ∈ Cor(U).

For any ~x ∈ R2 and k ∈ Z denote

Ak(~x) = bk~xcwhere the floor is taken for each coordinate separately:

b(x, y)c = (bxc, byc).Denote

Bk(~x) = Ak(~x)−Ak−1(~x) = bk~xc − b(k − 1)~xc.It easily follows that if ~x ∈ U then

Bk(~x) ∈ Cor(U).

Vector ~x will be represented as the two-way infinite sequence

. . . B−2(~x), B−1(~x), B0(~x), B1(~x), B2(~x), . . .

of corners. It is the balanced (or sturmian) representation of ~x.

Ak(~x) = bk~xc,Bk(~x) = Ak(~x)−Ak−1(~x) = bk~xc − b(k − 1)~xc.

The tile set corresponding to a rational affine map

fi(~x) = M~x +~b

and its domain square Ui consists of all tiles

fi(Ak−1(~x))

−Ak−1(fi(~x))

+(k − 1)~b

fi(Ak(~x))

−Ak(fi(~x))

+k~b

Bk(fi(~x))

Bk(~x)

where k ∈ Z and ~x ∈ Ui.

Ak(~x) = bk~xc,Bk(~x) = Ak(~x)−Ak−1(~x) = bk~xc − b(k − 1)~xc,fi(~x) = M~x +~b.

fi(Ak−1(~x))

−Ak−1(fi(~x))

+(k − 1)~b

fi(Ak(~x))

−Ak(fi(~x))

+k~b

Bk(fi(~x))

Bk(~x)

(1) For fixed ~x ∈ Ui the tiles for consecutive k ∈ Z match so that a

horizontal row can be formed whose top and bottom labels read the

balanced representations of ~x and fi(~x), respectively.


fi(Ak−1(~x))

−Ak−1(fi(~x))

+(k − 1)~b

fi(Ak(~x))

−Ak(fi(~x))

+k~b

Bk(fi(~x))

Bk(~x)

(2) A direct calculation shows that the tile computes function fi,

that is,

fi(~n) + ~w = ~s + ~e.


fi(Ak−1(~x))

−Ak−1(fi(~x))

+(k − 1)~b

fi(Ak(~x))

−Ak(fi(~x))

+k~b

Bk(fi(~x))

Bk(~x)

(3) Because fi is rational, there are only finitely many such tiles

(even though there are infinitely many k ∈ Z and ~x ∈ Ui). The tiles

can be effectively constructed.

If there is an infinite orbit then a tiling exists where the labels of

the horizontal rows read the balanced representations of the points

of the orbit:

Balanced representation of f(x)

Balanced representation of x



of the orbit:

2


Balanced representation of f (x)



of the orbit:

3Balanced representation of f (x)




of the orbit:


Balanced representation of f (x)4

Conclusion: the tile set we constructed admits a tiling of the plane

if and only if the system of affine maps is immortal. Undecidability

of the tiling problem follows from the undecidability of the

immortality problem.

The hyperbolic plane

The technique works well also in the hyperbolic plane. Hyperbolic

plane is a plane where Euclid’s fifth axiom does not hold: For any

point P and a line L that does not contain P there are more than

one lines through P that do not intersect L.

The hyperbolic plane

The technique works well also in the hyperbolic plane. Hyperbolic

plane is a plane where Euclid’s fifth axiom does not hold: For any

point P and a line L that does not contain P there are more than

one lines through P that do not intersect L.

To display hyperbolic geometry on the screen (=Euclidean plane)

we use the half-plane projection. The hyperbolic plane is

represented as the Euclidean half plane. The division line is the

horizon.

• Hyperbolic points are points in the open Euclidean half plane,

and

• hyperbolic lines are semi-circles whose centers are on the

horizon (and half-lines that are perpendicular to the horizon.)

The role of the Euclidean Wang square tile will be played by a

hyperbolic pentagon.

The pentagons can tile a ”horizontal row”.

”Beneath” each pentagon fits two identical pentagons. The

pentagons are all congruent (=isometric copies of each other), but

the projection makes objects close to the horizon seem smaller.

Infinitely many ”horizontal rows” fill the lower part of the half

plane.

Similarily the upper part can be filled. We see that the pentagons

tile the hyperbolic plane (in an uncountable number of different

ways, in fact.)

On the hyperbolic plane Wang tiles are pentagons with colorededges. Such pentagons may be placed adjacent if the edge colorsmatch. A given set of pentagons tiles the hyperbolic plane if atiling exists where the color constraint is everywhere satisfied.

The two sample tiles admit a tiling.

The hyperbolic tiling problem asks whether a given finite collection

of colored pentagons admits a valid tiling.

Theorem. The tiling problem of the hyperbolic plane is

undecidable.

Note that the hyperbolic Wang tiles can be transformed into

equivalent shapes exactly as in the Euclidean case: by introducing

different bumps and dents for different colors. So the undecidability

holds for the tiling problem using hyperbolic polygons.

We say that pentagon

r

n

ew

l

computes the affine transformation f : R2 −→ R2 if

f(~n) + ~w =~l + ~r

2+ ~e.

(Difference to Euclidean Wang tiles: The ”output” is now divided

between ~l and ~r.)

s

w e

Average = n

Average =

In a horizontal segment of length n where all tiles compute the

same f holds

f(~n) +1n

~w = ~s +1n

~e,


As the segment is made longer, the effect of the carry in and out

labels ~w and ~e vanish.

For a given system of affine maps fi and unit squares Ui we

construct for each i a set Ti of pentagons

• that compute function fi, and

• whose top edge labels ~n are in Ui.

It follows, exactly as in the Euclidean case, that if a valid tiling of

the hyperbolic plane with such pentagons exists then from the

labels of horizontal rows one obtains an infinite orbit in the system

of affine maps.

We still have to detail how to choose the tiles so that the converseis also true: if an immortal point exists then its orbit provides avalid tiling.

The tile set corresponding to a rational affine map

fi(~x) = M~x +~b

and its domain square Ui consists of all tiles

fi(Ak−1(~x))

− 12A2(k−1)(fi(~x))

+(k − 1)~b

fi(Ak(~x))

− 12A2k(fi(~x))

+k~b

B2k−1(fi(~x)) B2k(fi(~x))

Bk(~x)

where k ∈ Z and ~x ∈ Ui.


fi(Ak−1(~x))

− 12A2(k−1)(fi(~x))

+(k − 1)~b

fi(Ak(~x))

− 12A2k(fi(~x))

+k~b


Bk(~x)

(1) For fixed ~x ∈ Ui the tiles for consecutive k ∈ Z match so that a

horizontal row can be formed whose top and bottom labels read the

balanced representations of ~x and fi(~x), respectively.


fi(Ak−1(~x))

− 12A2(k−1)(fi(~x))

+(k − 1)~b

fi(Ak(~x))

− 12A2k(fi(~x))

+k~b


Bk(~x)

(2) A direct calculation shows that the tile computes function fi:

fi(~n) + ~w =~l + ~r

2+ ~e.


fi(Ak−1(~x))

− 12A2(k−1)(fi(~x))

+(k − 1)~b

fi(Ak(~x))

− 12A2k(fi(~x))

+k~b


Bk(~x)

(3) There are only finitely many such tiles (when fi is rational),

and they can be effectively constructed.

The tiles constructed admit a valid tiling iff the system of affine

maps has an immortal point:

Balanced representation of f(x)

Balanced representation of xBalanced representation of x

ConclusionA new proof for the undecidability of the tiling problem was

presented. The proof was based on a reduction where one

constructed tiles such that valid tilings are forced to simulate

iterations of a system of affine transformations.

The construction works well also in other set-ups. In particular, we

showed that the tiling problem in the hyperbolic plane is

undecidable.

jarkko kari- the tiling problem revisited

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